Question

Use a computer algebra system to find \(\mathbf{u} \times \mathbf{v}\) and a unit vector orthogonal to \(\mathbf{u}\) and \(\mathbf{v}\). $$ \begin{array}{l} \mathbf{u}=\langle-8,-6,4\rangle \\ \mathbf{v}=\langle 10,-12,-2\rangle \end{array} $$

Short answer

The cross product of vectors \( \mathbf{u} \) and \( \mathbf{v} \) is \( \langle 12, 4, 120 \rangle \) and the unit vector orthogonal to both these vectors is \( \langle 0.1, 0.03, 1 \rangle \) (to 2 decimal places).

Step-by-step solution

Compute the cross product

Firstly compute the cross product of vectors \( \mathbf{u} \) and \( \mathbf{v} \). You can do this by setting up a determinant where the first row is the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), the second row is the vector \( \mathbf{u} = \langle -8, -6, 4 \rangle \), and the third row is the vector \( \mathbf{v} = \langle 10, -12, -2 \rangle \). By expanding this determinant, you get the new vector which is the cross product.

Find the cross product

Calculate the cross product using the determinant formula. We get \[ \mathbf{u} \times \mathbf{v} = (-6 \times -2) - (4 \times -12) \mathbf{i} - ((-8) \times -2 - 10 \times 4) \mathbf{j} + ((-8) \times -12 - 10 \times (-6)) \mathbf{k} = \langle 12, 4, 120 \rangle \]. So, the cross product \( \mathbf{u} \times \mathbf{v} = \langle 12, 4, 120 \rangle \).

Normalize the Vector

To find the unit vector orthogonal to both \( \mathbf{u} \) and \( \mathbf{v}, \) take the vector \( \mathbf{u} \times \mathbf{v} \) and divide it by its magnitude. The magnitude of vector \( \mathbf{u} \times \mathbf{v} = \sqrt{(12)^2+(4)^2+(120)^2} = \sqrt{144+16+14400} = \sqrt{14560} = 120.67 \approx 120.7 (to 1 decimal place). Thus the unit vector is \( \langle 12/120.7, 4/120.7, 120/120.7 \rangle = \langle 0.099 \approx 0.1, 0.033 \approx 0.03, 0.994 \approx 1 \rangle \) (to 2 decimal places).